Choosing Thresholds in Markov Bound Problems

Introduction

Markov’s inequality turns the mean of a nonnegative random variable into a bound on its upper tail. The calculation is short; the careful part is identifying exactly which random variable is nonnegative and what threshold describes the event of interest. Rescaling, adding a constant, or applying a decreasing transformation can change that threshold—and sometimes change whether Markov’s inequality applies at all.

The guiding rule is to translate the event before calculating. Write the event as an upper-tail event for a nonnegative random variable, identify its positive threshold, and only then substitute the expectation and threshold into the inequality.

The Basic Inequality and Its Threshold

Theorem (Markov’s inequality): If $Y$ is a nonnegative random variable with $E[Y]<\infty$, then for every $t>0$, $$ P(Y\geq t)\leq \frac{E[Y]}{t}. $$ The same bound holds for $P(Y>t)$, since $\{Y>t\}\subseteq\{Y\geq t\}$.

The threshold $t$ must be positive. The numerator is the expectation of the same random variable $Y$ that appears in the event. These two pieces cannot be chosen separately: replacing $Y$ with a transformed variable requires translating the event to a threshold for that transformed variable.

Event-transfer principle: Suppose $Y\geq 0$ almost surely, $E[Y]<\infty$, and an event $A$ satisfies $A\subseteq\{Y\geq t\}$ for some $t>0$. Then $$ P(A)\leq \frac{E[Y]}{t}. $$

Proof. The event inclusion gives $P(A)\leq P(Y\geq t)$. Markov’s inequality applies because $Y$ is nonnegative, integrable, and $t>0$. Therefore $$ P(A)\leq P(Y\geq t)\leq \frac{E[Y]}{t}. $$ $\square$

This principle is useful when the event of interest is not initially written as a tail event for the variable whose mean is known. A transformation is helpful only if it supplies a valid nonnegative $Y$ and a positive threshold $t$.

Rescaling and Shifting: Track Both Parts

Let $X\geq 0$ almost surely with $E[X]<\infty$. For a positive constant $c$, the event $cX\geq u$, where $u>0$, is exactly $X\geq u/c$. Applying Markov’s inequality to either representation gives the same bound:

P(cX\geq u) = P\left(X\geq \frac{u}{c}\right) \leq \frac{E[X]}{u/c} = \frac{cE[X]}{u} = \frac{E[cX]}{u}.

The rescaling changes both the expectation and the threshold by the same factor. If the threshold is rescaled in the event but not in the inequality, the resulting calculation is wrong.

A shift also changes the threshold. If $Y=X+d$ is nonnegative almost surely, $E[|Y|]<\infty$, and $u>0$, then $$ P(X+d\geq u)=P(Y\geq u)\leq \frac{E[X]+d}{u}. $$ The condition $Y\geq 0$ must be checked; a shift does not automatically make an arbitrary random variable nonnegative. Moreover, if $X$ itself is already nonnegative, applying Markov directly to $X$ can give a tighter bound than applying it to $X+d$.

Worked Example: Comparing a Variable With Its Shifted Version

Suppose $X\geq 0$ almost surely and $E[X]=4$. Bound $P(2X+3\geq 15)$ in two ways.

First translate the event. Since $2$ is positive, $$ 2X+3\geq 15 \quad\Longleftrightarrow\quad 2X\geq 12 \quad\Longleftrightarrow\quad X\geq 6. $$ Markov’s inequality applied to $X$ gives $$ P(2X+3\geq 15)=P(X\geq 6)\leq \frac{E[X]}{6}=\frac{4}{6}=\frac{2}{3}. $$

Now apply it to the transformed variable. Set $Y=2X+3$. Since $X\geq 0$, $Y\geq 3$, so $Y$ is nonnegative. Also $E[Y]=2E[X]+3=11$. The event is $Y\geq 15$, and Markov gives $$ P(Y\geq 15)\leq \frac{E[Y]}{15}=\frac{11}{15}. $$ Both bounds are valid, but $2/3=10/15<11/15$. Here the shift raises the mean without raising the event threshold enough to improve the bound, so applying Markov directly to $X$ is better.

Monotone Transformations and Tail Direction

An increasing transformation preserves the direction of an upper-tail event: if $g$ is increasing and $g(x_0)=t$, then $X\geq x_0$ implies $g(X)\geq t$. A decreasing transformation reverses the direction. In particular, on positive values, taking reciprocals converts a lower-tail event into an upper-tail event: $$ X\leq a \quad\Longleftrightarrow\quad \frac{1}{X}\geq \frac{1}{a}, \qquad a>0. $$ This permits a Markov bound on a lower tail of $X$, provided $1/X$ is nonnegative and has finite expectation.

Threshold-translation rule: To use a transformed variable $Y=g(X)$, first establish an event inclusion $A\subseteq\{g(X)\geq t\}$ with $t>0$. Then check that $g(X)\geq 0$ almost surely and $E[g(X)]<\infty$. The resulting bound is $P(A)\leq E[g(X)]/t$.

Worked Example: Bounding a Lower Tail With a Reciprocal

Let $X$ equal $2$ or $10$, each with probability $1/2$. Find a Markov bound for $P(X\leq 3)$ using a reciprocal transformation, and compare it with the actual probability.

On the support of $X$, $X>0$. The event translates as $$ \{X\leq 3\}=\left\{\frac{1}{X}\geq \frac{1}{3}\right\}, $$ because taking reciprocals reverses the inequality for positive numbers. The transformed variable $Y=1/X$ is nonnegative and has finite mean: $$ E[Y]=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{10}\right) =\frac{1}{2}\left(\frac{5}{10}+\frac{1}{10}\right) =\frac{3}{10}. $$ The threshold is $1/3>0$, so Markov’s inequality gives $$ P(X\leq 3) \leq \frac{E[1/X]}{1/3} = \frac{3/10}{1/3} = \frac{9}{10}. $$ The actual probability is $P(X=2)=1/2$, which is indeed at most $9/10$. The bound is valid but not especially tight; a transformation makes a Markov bound possible here, but it does not guarantee a useful bound.

Worked Example: A Shift Makes Markov Applicable

Suppose $X=-2$ or $X=4$, each with probability $1/2$. Use $Y=X+5$ to bound $P(X+5\geq 8)$, and check the result against the actual probability.

The values of $Y$ are $3$ and $9$, so $Y\geq 0$ almost surely. Its expectation is $$ E[Y]=\frac{1}{2}(3)+\frac{1}{2}(9)=6. $$ The event is already expressed as $Y\geq 8$, with positive threshold $8$. Thus $$ P(X+5\geq 8)=P(Y\geq 8)\leq \frac{E[Y]}{8}=\frac{6}{8}=\frac{3}{4}. $$ In fact, $X+5\geq 8$ holds exactly when $X=4$, so the probability is $1/2$. The bound is correct. Applying Markov directly to $X$ would not be justified because $X$ takes the negative value $-2$.

Choosing and Checking the Bound

When several transformations are available, compare their resulting bounds rather than assuming a more elaborate transformation must be better. A candidate $Y$ must be nonnegative almost surely, have finite expectation, and represent the event as an upper tail at a positive threshold. Among valid choices, the bound with the smaller ratio $E[Y]/t$ is numerically more informative.

Write the target event.
Keep its original inequality and identify whether it is an upper tail, lower tail, or transformed event.

Choose a candidate variable.
Use the original variable if it is nonnegative; otherwise consider a transformation that is nonnegative on the support.

Translate the event exactly.
Find a positive $t$ for which the target event is contained in $\{Y\geq t\}$.

Check the hypotheses and calculate.
Verify $Y\geq 0$ almost surely and $E[Y]<\infty$, then evaluate $E[Y]/t$.

Common Mistakes

Keeping the old threshold after rescaling. From $cX\geq u$, for $c>0$, the threshold for $X$ is $u/c$, not $u$. Equivalently, applying Markov to $cX$ uses numerator $cE[X]$ and threshold $u$.
Applying Markov to a variable that can be negative. A finite expectation alone is not enough. The variable in the inequality must be nonnegative almost surely.
Using a nonpositive transformed threshold. The bound $E[Y]/t$ requires $t>0$. A threshold of zero or less does not give the stated Markov bound.
Reversing a decreasing transformation incorrectly. For $X>0$, $X\leq a$ implies $1/X\geq 1/a$, not $1/X\leq 1/a$.
Assuming every valid bound is informative. A bound greater than $1$ is still formally valid, but the basic probability bound $P(A)\leq 1$ is stronger.

Key takeaway: Markov’s inequality is applied to a nonnegative variable at the threshold that describes the transformed event. Correctly translating the event—and checking nonnegativity and a positive threshold—comes before substituting an expectation.

Check Your Understanding

For each problem, identify the nonnegative variable and its threshold before computing a bound.

If $X\geq 0$ and $E[X]=5$, what Markov bound follows for $P(3X\geq 30)$? Show the equivalent threshold for $X$.
If $X\geq -4$ and $E[X]=2$, use $Y=X+4$ to bound $P(X\geq 6)$. State why Markov applies to $Y$.
For positive $X$, translate the event $X\leq a$, where $a>0$, into an upper-tail event for $1/X$. What additional expectation condition is needed to apply Markov?
Explain why $P(X\geq 5)\leq E[X]/5$ is not justified solely by knowing that $E[X]$ is finite.
If a valid transformed-variable calculation gives a bound of $1.2$, what conclusion can you report as the more informative probability bound?